Let $f$ be a twice differentiable function such that $f''\left( x \right) = - f\left( x \right)$ and $f'\left( x \right) = {\rm{g}}\left( x \right).$ If $h'\left( x \right) = \left[ {f{{\left( x \right)}^2} + {\rm{g}}{{\left( x \right)}^2}} \right]h\left( 1 \right) = 8$ and $h\left( 0 \right) = 2,$ then $h\left( 2 \right)$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

Let

f

be a twice differentiable function such that

f''\left( x \right) = - f\left( x \right)

and

f'\left( x \right) = \mathrm{g}\left( x \right).

h'\left( x \right) = \left[ {f{{\left( x \right)}^2} + \mathrm{g}{{\left( x \right)}^2}} \right]h\left( 1 \right) = 8

and

h\left( 0 \right) = 2,

then

h\left( 2 \right)

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because after solving the differential equation and using the given conditions, we find that

h(2) = 14

, which is not among the provided options.